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I want to find the total impendance of a coil. I think I know how to do it but I'm not sure.Z X + RX 2(pi)wLL (uAN^2)/lZ is impendance, R is resistance of the wire of the coil, X is the reactance, w is current frequency in hertz, L is inductance, u is permeability of the core, A is area of the coil loops, N is number of turns in coil, and l is the length of the coil.Together this would yieldZ (2(pi)wuAN^2)/l + RWould this be correct?: All these formulas only apply to a sine wave single frequency. Z is from ohms law if you have measured V and I: Z + V/I Ideally Z is stated as a vector |Z|, which means a magnitude in ohms and a phase angle in degrees. If X and R are known: Z + sqrt(R^2 + X^2). This is using trigonometry to account for the resulting direction of the vector (phase angle) of the two components R and X. Some calculations involving power, voltage, impedance and current are easier using vectors and trigonometry, while others are easier using complex numbers (real and imaginary component). Complex numbers have a set of rules for arithmetic using them. Adding them as for two impedances in series is very straight forward. Z as a complex number is R + jX The following formulas allow for converting from one representation to the other. PF (Power factor) true power / apparent power cos(θ) from trig. θ arccos(PF) |Z| V / I with a phase angle (θ) Rseries |Z| cos(θ) Xseries |Z| sin(θ) Z R + jX |Z| sqrt(R^2 + X^2) where X is the net reactance of XL and Xc. θ arctan(Xs / Rs)s

A 24.0 V battery is in series with a switch, a 0.40 H inductor, time constant of 16 ms, and a 25 ohm resistor.(a) Find the potential differences across the inductor at t τ. V(1L) (b) Find the potential differences across the resistor at t τ. V(1R) (c) Find the potential differences across the inductor at t 3τ. V(2L) (d) Find the potential differences across the resistor at t 3τ. V(2R) Me and my partner are completely stuck on here, if someone could help i would be incredibly greatful!: by ability of Ohms regulation, the Viktage throughout the time of a resistor is the present in it cases the resistance. Voltage of Resistor equals (3.0)(3.5)10.5 V. Voltage throughout the time of an inductor is the inductance cases the cost of replace of the present. Voltage of Inductor equals (2.8)(.260).728 V. including those mutually provides 11.2 Vs

are connected to 240V 50 Hz AC mains. a current of 3A flows lagging 31degrees behind the supply of voltage while the voltage across the inductor is 171V.determine the resistance and inductive reactance.: If I told you how long ago it was when I last did problems like this you wouldn`t believe me! However I am tempted to have a go. For calculation purposes , let 3A current lag by 30 ° rather than the 31 ° you suggest. Also let voltage across inductor be 170 V Vr ? 240 ? - 170 ? Vr 170 V (voltage across resistor) 170/3 R 57? let inductive reactance X ? Voltage across inductance 170 v I 170 / X 3X 170 X 57? I will be pleasantly surprised if this is correct !s

I always thought that capacitor or inductor circuits invariably contain a voltage thats 90 degrees phase shifted off the ccurrent, however yesterday I was reading some material and I came across something that states that capacitors or inductors can phase shift the voltage by 45 degrees by altering to the capacitance or inductance appropriately.here is what I read:On starting, the switch is closed, placing the capacitor in series with the auxiliary winding. The capacitor is of such a value that the auxiliary winding is effectively a resistive-capacitive circuit in which the current leads the line voltage by approximately 45°. The main winding has enough inductance to cause the current to lag the line voltage by approximately 45°. The two currents are therefore 90° out of phase, and so are the magnetic fields which they generate. The effect is that the two windings act like a two-phase stator and produce the revolving field required to start the motor.Can someone account for this?Thanks in advance: The circuit could give a 45? phase shift since it contains capacitance, inductance and resistance (from the windings). A vector diagram for this arrangement would show: VR horizontal VL vertical (up) VC vertical (down) The resultant could be any angle between 0 and 90?, depending on the values of VR, VL and VC.s

here's the text which contains it,Referring to figure, it can be seen that no power is disspated in a pure inductor. In the first quarter of cycle, both V and I are positive so the power is positive, which means that energy is supplied to the inductor. In the second quarter, V is positive but I is negative. Now power is negative which implies that the energy is returned by the inductor(The figure looks something like this) ::: Yes, you're correct. Your picture shows the correct phase relationship between voltage and current in an inductor, and in that case your description is correct.s

A probe is attached to the Infrared LED and use the adjustable resistor R1 to calibrate the emitter to transmit a 5 kHz square wave.Data: The emitter circuit on the left produces light pulses at a frequency determined by the variable resistor R1. That resistor is adjusted to tune the frequency to the frequency peak of the narrow-band amplifier of the receiver on the right. That frequency is fixed by the capacitor-inductor tuning circuit (near ICA2).By other definitions: The variable resistor R3 near ICA2 across the LC circuit flattens the respons of the LC circuit and increases the IC2A filter respons bandwidth.So does that mean that I use the variable resistor R3 near ICA2 to modulate a frequency of 5kHz?If not, how is the 5kHz created in the receiver circuit and, what am I suposed to adjust the R3 according to? Is it suposed to be. R3 sets the overal sensitivity but what voltage or something else do I adjust it to?: First R1 is adjusted so that the LED puts out pulses at 5 kilohertz. f 1.44/((2*R1 + 10,000)(0.0047 uF)). where f is 5,000 hertz (2*R1 + 10,000)(0.0047 uF) 2.88 x 10^-4 seconds which is 1.44/5,000 hertz R1*0.0094u sec + 4.7 x 10^-5 2.88 x 10^-4 seconds R1* 9.4 x 10^-9 2.41 x 10^-4 seconds R1 is about 25.6 kohms, this sets the transmit frequency The 100 k ohm resistor across the LC tank circuit is used to flatten the response. It DOESN'T modulate the 5 kilohertz frequency. It appears to be optional and is most likely not needed. Try the circuit without R3, R3 will reduce the peak response of the tank circuit. If the circuit doesn't ever seem to trip when the beam is broken, it may need to be incorporated. Use a linear not log pot and set it to the halfway point i.e. 50 kohm The NPN photo transistor conducts when the LED is emitting photons. The current pulses from the photo transistor will be 5 kilohertz pulses i.e. will occur every 200 microseconds. I would leave R3 out and see how the circuit responds. As long as LC satisfies this equation the circuit should work fine f 1/(2*π*√(L*C)) where f is 5,000 hertz √(LC) 1/(6.28*5,000) 31.8 micro seconds LC 1 x 10^-9 s^2 C 1 x 10^-7 farads L 1 x 10^-2 Henry L should be 10 milli Henry not 10 micro Henry So change L to 10 mH, the circuit will not work with a 10 uH inductor.s

A certain circuit consists of an inductor of 55 mH in series with a resistor of 90 Ω. At a moment when the current in the circuit is 21 A, a switch in the circuit is opened. How long will it take for the current to fall to 7.14 A? Answer in units of s.Okay, easy smeasy I thought. i made an expression 7.1421(1-e^(-90t/55*10^-3))Simply algebra, so I got t2.53926x10^-4. This is wrong! please show me the errors of my ways. Do the entire problem. Show work. I need to get this down to I can tackle the tougher problems.And I based my expression off of the equation i(Io)(1-e^(rt/L)). Thanks! will choose best answer!: This is easy. You used the wrong formula. The formula is for current INCREASING from zero and asymptoting to Io. But it isn't accurate even for that. The actual formula is i Io (e ^- t/(rL)) Note the features. The larger the value of r the less that is lost (or gained ) per second. The larger the value of L the less that is lost (or gained) per second so these must be in the denominator. The larger the time the more that is lost so it must be in the numerator. The negative sign in e ^ - x means that as x becomes larger ( more time taken) the answer becomes smaller. If we had 1 - e^ -x then it would asymptote to 1 not to zero. This would be the case for increasing the current to a maximum.s

Calculate the inductance of an air-core coil with the following specifications: length 20 cm, #of turns 200 and diameter of core 2 cm.Could you show your work if you know how to do it? Thanks :): See: okorders

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